Question

Water at 60 °F flows through the elbow as shown below and is then injected to the atmosphere through a nozzle (on a horizontal plane). The pipe diameter is Di = 4 in. while the diameter of the exit of the nozzle is D2 = 1 in. At a flow rate of Q = 245 gpm, the gage pressure at the section (1) where the flange locates is Pi = 34 psig. Neglect the weight of the water and elbow, you are asked to do the following: 1.1 Calculate the mean water velocity at section 1 (Vi) and at the exit of the nozzle (V2). 1.2 Determine the force on the flange bolts at section 1. 1.3 Determine the total head loss of the elbow and the nozzle system, hl. 40°

Answer 1

1 40 16) 2 Grivon Pipe diameter (D) = 4 in. Nozzle Exit diometer (dl = 1in. flow Rote = 245gbm. Pressure At 1 P, = 34 P, = 34 Psig. goge solution- PARICA) density of Water At bof = 1.9383 slug ft² 0 = 2 45 gom 0.5459 ft /see 1 us gollon = 3.7854 litrs. 1 ft = 12 inch. 34 Psia = 4896 4896 al ft²

2 o = A x V 0.5459 - D, XVI T 4 0.5459 1 x ( 4 ) xv, V = 6.255 ft/see velocity (meon) Section 1 vi = 6.255 ft see Apply continuity Equation At Nozzle Entry And Exit A v A2 V2 II (4 3 x 6.255 6.255 - :(1+2 X V2 V₂ = 100.08 ft /see. PART-B Lincor Momentum Apply conservation between 1 and 2 Momentum conservation in x x - direction. (P. A + P2 A2 cos46 ) + x = (im u Gos40-nv.)

Pr is At monspheric so goge Pressure (P, A, & fx) = (-mv, cos 4o-inv.) 48964 + fx = - 80/100.08x cosuo 6.255) fx = -1.9383& 0.5459x70.4002 427.256 + CL Fx = - 501.758 lbf. Momentum conservation in Y direction (o + P2 A2 sin 40 ) + fy = n t'ksin4o-o) P2 Aqsinuo & fy = - insinuo P = 0 because Atmonspheric Pressure) fy = – im ve sin 40 fy = - 1.29383 x 0.5459 x 100.04 x sinuo - - 68.04 fy Ibf.

So foru Acting on Bolt f = √ (F x + (fy f = (501.758) + (68.04) 2 F = 506.3500 lbf F 11 506-350.4 lbf PART (6) Apply Energy Equation between and Nozzle Exit. Pitutt la + color + Lat nu s 9 2g 2 2 4896 (6.255) (100.05) 11 + hi 2 x 32.2 1.9383 x 32-2 2x 32.2 ha = 76.38 ft